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\title{\heiti\zihao{2} 复变函数-第7章}
\author{20373963-樊若宸}
\date{\today}

\begin{document}
\maketitle
\section{试求 $f(t)=|\sin t|$ 的离散频谱和它的傅里叶级数的复指数形式.}
\textbf{解:}\quad






\section{试证:若 $f(t)$ 满足傅里叶积分定理的条件, 则有$$f(t)=\int_{0}^{+\infty} a(\omega) \cos \omega t \mathrm{~d} \omega+\int_{0}^{+\infty} b(\omega) \sin \omega t \mathrm{~d} \omega$$其中$$\begin{array}{l}a(\omega)=\dfrac{1}{\pi} \int_{-\infty}^{+\infty} f(\tau) \cos \omega \tau \mathrm{d} \tau \\b(\omega)=\dfrac{1}{\pi} \int_{-\infty}^{+\infty} f(\tau) \sin \omega \tau \mathrm{d} \tau\end{array}$$}
\begin{proof}
    
\end{proof}


\section{求下列函数的傅里叶变换.}
(1)矩形脉冲函数
$$
f(t)=\left\{\begin{array}{ll}
A, & 0<t<\tau, A \neq 0 \\
0, & \text { 其他 }
\end{array}\right.
$$
\textbf{解:}\quad
$$
\begin{aligned}
    F(\omega) &= \int_{-\infty}^{+\infty}f(t)\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\int_0^{\tau}A\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\left.A\dfrac{\mathrm{e}^{-i\omega t}}{-i\omega}\right|_{0}^{\tau}\\
    &=\dfrac{A(1-\mathrm{e}^{-i\omega\tau})}{i\omega}
\end{aligned}
$$

(2)函数
$$
f(t)=\left\{\begin{array}{ll}
1-t^{2}, & |t| \leqslant 1 \\
0, & |t|>1
\end{array}\right.
$$
\textbf{解:}\quad
$$
\begin{aligned}
    F(\omega) &= \int_{-\infty}^{\infty}f(t)\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\int_{-1}^1(1-t^2)\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\left.(1-t^2)\dfrac{\mathrm{e}^{-i\omega t}}{-i\omega}\right|_{-1}^{1}+\int_{-1}^{1}2t\dfrac{\mathrm{e}^{-i\omega t}}{-i\omega}\mathrm{d}t\\
    &=\int_{-1}^{1}2t\dfrac{\mathrm{e}^{-i\omega t}}{-i\omega}\mathrm{d}t\\
    &=\left.2t\dfrac{\mathrm{e}^{-i\omega t}}{-\omega^2}\right|_{-1}^{1} - \int_{-1}^{1}2\dfrac{\mathrm{e}^{-i\omega t}}{-\omega^2}\mathrm{d}t\\
    &=-2\dfrac{\mathrm{e}^{i\omega}+\mathrm{e}^{-i\omega}}{\omega^2} + \dfrac{2}{\omega^2}\left.\dfrac{\mathrm{e}^{-i\omega t}}{-i\omega}\right|_{-1}^{1}\\
    &=\dfrac{4\sin w - 4 \omega \cos \omega}{\omega^3}
\end{aligned}
$$

\section{求下列函数的傅里叶积分.}
(1)
$$
f(t)=\left\{\begin{array}{ll}
0, & t<0 \\
\mathrm{e}^{-t} \sin 2 t, & t \geqslant 0
\end{array}\right.
$$
\textbf{解:}\quad
函数的Fourier变换为
$$
\begin{aligned}
    F(\omega) &= \int_0^{+\infty}\mathrm{e}^{-t}\sin 2t\mathrm{e}^{i\omega t}\mathrm{d}t\\
    &=\int_0^{+\infty}\sin 2t\mathrm{e}^{-(1+i\omega) t}\mathrm{d}t\\
    &=-\dfrac{1}{2}\left(\left.\cos 2t\mathrm{e}^{-(1+i\omega)t}\right|_{0}^{+\infty}+(1+i\omega)\int_{0}^{\infty}\cos 2t\mathrm{e}^{-(1+i\omega)t}\mathrm{d}t\right)\\
    &=\dfrac{1}{2}-\dfrac{1+i\omega}{4}\left[\left.\sin 2t\mathrm{e}^{-(1+i\omega)t}\right|_{0}^{+\infty}+(1+i\omega)\int_0^{+\infty}\sin 2t\mathrm{e}^{-(1+i\omega)t}\mathrm{d}t\right]\\
    &=\dfrac{1}{2}-\dfrac{(1+i\omega)^2}{4}F(\omega)
\end{aligned}
$$

从而
$$
F(\omega) = \dfrac{2}{4+(1+i\omega)^2}
$$

Fourier积分为
$$
\begin{aligned}
    f(t) = \dfrac{1}{\pi}\int_{-\infty}^{+\infty}\dfrac{1}{4+(1+i\omega)^2}\mathrm{e}^{i\omega t}\mathrm{d}\omega
\end{aligned}
$$


(2)
$$
f(t)=\left\{\begin{array}{ll}
-1, & -1<t<0 \\
1, & 0 \leqslant t<1 \\
0, & \text { 其他 }
\end{array}\right.
$$
\textbf{解:}\quad
函数的Fourier变换为:
$$
\begin{aligned}
    F(\omega) &= \int_{-1}^0-\mathrm{e}^{-i\omega t}\mathrm{d}t + \int_0^{1}\mathrm{e}^{-i\omega t}\mathrm{d}t\\
    &=\dfrac{2 - 2\cos \omega}{i\omega}
\end{aligned}
$$

Fourier积分为
$$
\begin{aligned}
    f(t) = \dfrac{1}{\pi}\int_{-\infty}^{\infty}\dfrac{1-\cos\omega}{i\omega}\mathrm{e}^{i\omega t}\mathrm{d}\omega
\end{aligned}
$$



\section{求下列函数的傅里叶变换, 并证明下列积分结果.}
(1) $f(t)=\mathrm{e}^{-\beta|t|}(\beta>0)$, 证明
$$
\int_{0}^{+\infty} \dfrac{\cos \omega t}{\beta^{2}+\omega^{2}} \mathrm{~d} \omega=\dfrac{\pi}{2 \beta} \mathrm{e}^{-\beta|t|}
$$
\begin{proof}
    函数的Fourier变换为
    $$
    \begin{aligned}
        F(\omega) &= \int_{-\infty}^{0}\mathrm{e}^{(\beta - i\omega)t}\mathrm{d}t+\int_0^{+\infty}\mathrm{e}^{-(\beta + i\omega)t}\mathrm{d}t\\
        &=\dfrac{1}{\beta - i\omega}+\dfrac{1}{\beta + i\omega}\\
        &=\dfrac{2\beta}{\beta^2+\omega^2}
    \end{aligned}
    $$
    函数的Fourier积分为
    $$
    \begin{aligned}
        f(t) &= \dfrac{1}{\pi}\int_{-\infty}^{+\infty}\dfrac{\beta}{\beta^2+\omega^2}\mathrm{e}^{i\omega t}\mathrm{d}\omega\\
        &=\dfrac{\beta}{\pi}\int_{-\infty}^{+\infty}\dfrac{\cos \omega t + i\sin \omega t}{\beta^2+\omega^2}\mathrm{d}\omega\\
        &=\mathrm{e}^{-\beta|t|}
    \end{aligned}
    $$
    从而因为其虚部为$0$,故
    $$
    \int_{-\infty}^{+\infty}\dfrac{ i\sin \omega t}{\beta^2+\omega^2}\mathrm{d}\omega = 0
    $$
    并且
    $$
    \int_{-\infty}^{+\infty}\dfrac{\cos \omega t }{\beta^2+\omega^2}\mathrm{d}\omega = 2\int_{0}^{+\infty}\dfrac{\cos \omega t }{\beta^2+\omega^2}\mathrm{d}\omega=\mathrm{e}^{-\beta|t|}
    $$
    从而
    $$
    \int_{0}^{+\infty}\dfrac{\cos \omega t }{\beta^2+\omega^2}\mathrm{d}\omega = \dfrac{\pi}{2\beta}\mathrm{e}^{-\beta|t|}
    $$
\end{proof}



(2)
$$
f(t)=\left\{\begin{array}{ll}
\sin t, & |t| \leqslant \pi \\
0, & |t|>\pi
\end{array},\right.
$$
证明
$$
\int_{0}^{+\infty} \dfrac{\sin \omega \pi \sin \omega t}{1-\omega^{2}} \mathrm{~d} \omega=\left\{\begin{array}{ll}
\dfrac{\pi}{2} \sin t, & |t| \leqslant \pi \\
0, & |t|>\pi
\end{array}\right.
$$


\end{document}